For what values of x, if any, does f(x) = cot((7pi)/12-x)  have vertical asymptotes?

Oct 31, 2017

$x = \frac{7 \pi}{12} - 2 \pi n$
$x = - \frac{5 \pi}{12} - 2 \pi n$

For $n \in \mathbb{Z}$

Explanation:

To approach this we must cosnider how $\cot x = \cos \frac{x}{\sin} x$

Hence for $\cot x$ to have verticle asymptotes, the denominator must be 0, so hence $\sin x = 0$

So hence in this circumstance ;
$\sin \left(\frac{7 \pi}{12} - x\right) = 0$ for the verticle asymptotes

So hence now we consider the general solution of;
$\sin x = \sin a$
Then $x = 2 \pi n + a$
and $x = 2 \pi n + \pi - a$

So hence $\sin \left(\frac{7 \pi}{12} - x\right) = \sin 0$

Hence; $\frac{7 \pi}{12} - x = 2 \pi n$
and $\frac{7 \pi}{12} - x = 2 \pi n + \pi$

Hence rearanging to yeild;

$x = \frac{7 \pi}{12} - 2 \pi n$
$x = - \frac{5 \pi}{12} - 2 \pi n$

For $n \in \mathbb{Z}$

We can consider this by graphing;
graph{cot( 7pi/12 -x) [-4.354, 5.644, -2.28, 2.72]}